Networking - Networking Basics - Discussion

Subnet should be C 192.168.168.128-190.

I think answer should be B. Because 2nd subnet should be 128-192. I. E 128 is network address and 192 is broadcast address. Hence valid host range should be 129-191. So answer is B.

I can't understand. Please anyone help me.

Here are the answers using 255.255.255.192:

How many subnet bits are used in this mask? Answer: 2 2^2-2=2 subnets.

How many host bits are available per subnet? Answer: 6 2^6-2=62 hosts per subnet.

What are the subnet addresses?Answer: 256-192=64 (the first subnet)64+64=128 (the second subnet)64+128=192. However, although 192 is the subnet mask value, it's not a valid subnet. The valid subnets are 64 and 128.

What is the broadcast address of each subnet?Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191.

What is the valid host range of each subnet? Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190.

Why other options are in correct, anyone can explain it briefly?

Thank you @Lavanya S S.

Hi, I am not getting this, Please explain it in detail.

Hello, all.

The answer is obtained as follows:

IP: 192.168.168.188
Subnet: 255.255.255.192

192 in binary :
1 1 0 0 0 0 0 0
188 in binary:
1 0 1 0 1 1 0 0

192 AND 188:
1 1 0 0 0 0 0 0
1 0 1 0 1 1 0 0
Ans: 1 0 0 0 0 0 0 0

In 192 first 2 digits are 1 therefore in Ans 1 0 will be part of subnet id.

While remaining 6 will give host address;

First Host addr: 1 0 0 0 0 0 0 1 = 129.
Last host addr: 1 0 1 1 1 1 1 0 = 190.

And Broad cast addr will be : 1 0 1 1 1 1 1 1.

How you get 192.168.168.191 as broadcast id?

It's simple guys.

Given IP address : 192.168.168.188.
Given subnet mask address:- 255.255.255.192.

To find the N/W ID
step1: Convert both addresses into binary form.
step2: Perform AND operation b/w them.
step3: Result will be your network ID i.e(192.168.168.128).
step4: So range of the N/W starts from 192.168.168.129 (starting host ID).
step5: The last id of the N/w is 192.168.168.191 (which is broadcast domain).
step6: So the last id of the N/W will be 192.168.168.190.

Thank you.